Partial Fraction Decomposition

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• Jul 12th 2010, 08:38 AM
MechEng
Partial Fraction Decomposition
Good afternoon,

I am really stuck on a particular partial fraction decomposition problem.

I have:

$\frac{x^2+x+1}{x^4+2x^2}$

Which easily becomes:

$\frac{x^2+x+1}{x^2(x^2+2)}$

Now, should this be setup like?:

$\frac{A}{x}+\frac{B}{x^2}+\frac{Cx+D}{x^2+2}$

If this should be setup like this, I am confused about where to turn next.
• Jul 12th 2010, 09:15 AM
Ackbeet
For any partial fractions setup, the next thing you need to do is set your setup equal to the original fraction, get common denominators, equate like powers of x, and solve for your constants. Make sense?
• Jul 12th 2010, 10:02 AM
MechEng
So, should that look like:

$\frac{x^2+x+1}{x^4+2x^2} = \frac{A(x^3+2x)}{x} + \frac{B(x^2+2)}{x^2} + \frac{(Cx+D)(x^2)}{x^2+2}$

Subsequently I get:

$x^2+x+1 = A(x^3+2x) + B(x^2+2) + (Cx+D)(x^2)$

Should I then set x=0 to get:

1 = 2B

$B=\frac{1}{2}$

Substituting B=1/2 back in, I get

$x^2+x+1 = A(x^3+2x) + \frac{(x^2+2)}{2} + (Cx+D)(x^2)$

Where do I go from here?
• Jul 12th 2010, 10:06 AM
Ackbeet
Everything looks fine so far. I wouldn't actually have thought of setting x=0, but that works well in quickly obtaining one of the constants. You need to equate the coefficients of like powers of x. For the $x^{2}$ powers, for example, you have $1=B+D$. You'll get a system of equations you'll have to solve simultaneously. Can you finish from here?
• Jul 12th 2010, 10:13 AM
MechEng
[QUOTE=Ackbeet;536165]You need to equate the coefficients of like powers of x. [QUOTE]

I'm not sure I follow you here. My book doesn't do a particularly great job of explaining this.
• Jul 12th 2010, 10:28 AM
Ackbeet
Ok, let me re-write your equation there: you have

$x^{2}+x+1 = A(x^{3}+2x) + B(x^{2}+2) + (Cx+D)(x^{2}).$

The LHS does not need to be altered, because it already looks the way I want. The RHS I alter such that you get:

$x^{2}+x+1 = (A+C)x^{3}+(B+D)x^{2}+(2A)x+(2B).$

Now, in order for this equation to hold, you have to equate the coefficients of like powers of x, or else one term will start growing too quickly. That is, you can read off the following four equations immediately:

$0=A+C$

$1=B+D$

$1=2A$

$1=2B$

Incidentally, the process of solving this system will tell you if you used the right initial guess for your partial fraction expansion. You should get one unique solution. Otherwise, you'll need to change your initial guess.
• Jul 12th 2010, 11:10 AM
MechEng
Quote:

Originally Posted by Ackbeet
That is, you can read off the following four equations immediately:

$0=A+C$

$1=B+D$

$1=2A$

$1=2B$

I think I'm missing something here, where did the values on the LHS come from?

Ah... I think I see what happened here...

Since there are no x^3's on the LHS, that eqn. is equal to zero. Since the x^2 on the LHS divided by the x^2 on the RHS equals 1, we have that eqn. equal to 1.

Am I following along correctly here?
• Jul 12th 2010, 11:38 AM
Ackbeet
What we're really doing here is reading off coefficients. If I were to give you the expression $9x^{2}-7$, answer the following questions:

1. What is the coefficient of the $x^{2}$ term?
2. What is the coefficient of the $x^{0}$ (constant) term?
3. What is the coefficient of the $x^{3}$ term?
4. What is the coefficient of the $x^{1}$ term?
• Jul 12th 2010, 11:41 AM
MechEng
1. 9
2. -7
3. 0
4. 0

Are we examing only the LHS or RHS (in your previous post) in this case?

Thank you very much for your patience with me.
• Jul 12th 2010, 11:44 AM
Ackbeet

Now, with the equations I gave you in post #6, we're looking at both sides, and equating the two. The coefficient of $x^{2}$ on the LHS must equal the coefficient of $x^{2}$ on the RHS. The coefficient of $x^{3}$ on the LHS must equal the coefficient of $x^{3}$ on the RHS. And so on.
• Jul 12th 2010, 11:54 AM
MechEng
Ah...

I have no idea why that wasn't clear from the beginning...

Thank you.
• Jul 12th 2010, 11:57 AM
Ackbeet
Hehe. I love the light-bulb moments! That's what tutors live for.

So, what do you get for your constants?
• Jul 12th 2010, 12:10 PM
MechEng
I get:

$\frac{x^2+x+1}{x^4+2x^2} = \frac{1}{2x} + \frac{1}{2x^2} + \frac{1-x}{2(x^2+2)}$

Or is that better expressed as?:

$\frac{x^2+x+1}{x^4+2x^2} = \frac{1}{2x} + \frac{1}{2x^2} + \frac{1}{2(x^2+2)} - \frac{x}{2(x^2+2)}$
• Jul 12th 2010, 12:12 PM
Ackbeet
I'd probably pick the first expression. Query: how can you double-check your answer?
• Jul 12th 2010, 12:45 PM
MechEng